Chapter 4.3, Problem 7PPSE

Solution

The equation of the parabola in standard form which passes through the points $\left(1,-2\right)\text{,}\left(2,-2\right)\text{ and }\left(3,-4\right)$ .

The equation of the parabola in standard form is $y=-x^2+3x-4$ .

Given:

The parabola passes through the points $\left(1,-2\right)\text{,}\left(2,-2\right)\text{ and }\left(3,-4\right)$ .

Concept used:

The equation of the parabola in standard form is $y=a{x}^2+bx+c,a\ne 0$ can be obtained by satisfying the given points in the parabola.

Calculation:

Consider the equation of the parabola in standard form is $y=a{x}^2+bx+c$ where $a\ne 0$ .

Since, the parabola passes through the points $\left(1,-2\right)\text{,}\left(2,-2\right)\text{ and }\left(3,-4\right)$ . Therefore, the three points $\left(1,-2\right)\text{,}\left(2,-2\right)\text{ and }\left(3,-4\right)$ will satisfy the parabola $y=a{x}^2+bx+c$ where $a\ne 0$ .

Satisfying the points $\left(1,-2\right)\text{,}\left(2,-2\right)\text{ and }\left(3,-4\right)$ in the parabola $y=a{x}^2+bx+c$ .

  $\begin{array}{c}y=a{x}^2+bx+c\mathrm{....}\left(\text{1}\right)\\ -2=a{\left(1\right)}^2+b\left(1\right)+c\mathrm{....}\left(\text{Satisfy point }\left(1,-2\right)\text{ in }\left(1\right)\right)\\ -2=a+b+c\mathrm{....}\left(2\right)\\ -2=a{\left(2\right)}^2+b\left(2\right)+c\mathrm{....}\left(\text{Satisfy point }\left(2,-2\right)\text{ in }\left(1\right)\right)\\ -2=4a+2b+c\mathrm{....}\left(3\right)\\ -4=a{\left(3\right)}^2+b\left(3\right)+c\mathrm{....}\left(\text{Satisfy point }\left(3,-4\right)\text{ in }\left(1\right)\right)\\ -4=9a+3b+c\\ -4=3\left(3a+b\right)+c\mathrm{....}\left(4\right)\end{array}$

Eliminate $\left(c\right)$ from $\left(2\right)\text{and }\left(3\right)$ gives $3a+b=0$ .

Substitute $3a+b=0$ in $\left(4\right)$ gives, $c=-4$ .

Substitute $c=-4$ in $\left(2\right)\text{and }\left(3\right)$ and simplifying gives,

  $\left.\begin{array}{c}a+b=2\\ 2a+b=1\end{array}\right\}\mathrm{....}\left(5\right)$

Solving the set of equations $\left(5\right)$ simultaneously gives, $a=-1\text{ and }b=3$ .

Now, plugging the parameters $a=-1,b=3\text{ and }c=-4$ in standard form of parabola $\left(1\right)$ gives the required equation of parabola in standard form $y=-x^2+3x-4$ .